PROBLEM 12.004
KNOWN: Furnace with prescribed aperture and emissive power.
FIND: (a) Position of gauge such that irradiation is G = 1000 W/m2, (b) Irradiation when gauge is tilted
θd = 20o, and (c) Compute and plot the gage irradiation, G, as a function of the separation distance, L, for
the range 100 ≤ L ≤ 300 mm and tilt angles of θd = 0, 20, and 60o.
SCHEMATIC:
ASSUMPTIONS: (1) Furnace aperture emits diffusely, (2) Ad << L2.
ANALYSIS: (a) The irradiation on the detector area is defined as the power incident on the surface per
unit area of the surface. That is
G = q f →d Ad
q f → d = Ie Af cos θ f ω d − f
(1,2)
where q f →d is the radiant power which leaves Af and is intercepted by Ad. From Eqs. 12.2 and 12.5,
ω d − f is the solid angle subtended by surface Ad with respect to Af,
ω d − f = Ad cos θ d L2 .
(3)
Noting that since the aperture emits diffusely, Ie = E/π (see Eq. 12.14), and hence
(
G = ( E π ) Af cos θ f Ad cos θ d L2
) Ad
(4)
Solving for L2 and substituting for the condition θf = 0o and θd = 0o,
L2 = E cos θ f cos θ d A f π G .
(5)
1/ 2
π

(20 × 10−3 ) 2 m 2 π × 1000 W m 2 
= 193 mm .
4

(b) When θd = 20o, qf→d will be reduced by a factor of cos θd since ωd-f is reduced by a factor cos θd.
Hence,


L = 3.72 × 105 W m 2 ×
<
<
G = 1000 W/m2 × cos θd = 1000W/m2 × cos 20o = 940 W/m2 .
(c) Using the IHT workspace with Eq. (4), G is computed and plotted as a function of L for selected θd.
Note that G decreases inversely as L2. As expected, G decreases with increasing θd and in the limit,
approaches zero as θd approaches 90o.
Irradiation, G (W/m^2)
3000
2000
1000
0
100
200
Separation distance, L (mm)
thetad = 0 deg
thetad = 20 deg
thetad = 60 deg
300
PROBLEM 12.17
KNOWN: Sun has equivalent blackbody temperature of 5800 K. Diameters of sun and earth as well
as separation distance are prescribed.
FIND: Temperature of the earth assuming the earth is black.
SCHEMATIC:
ASSUMPTIONS: (1) Sun and earth emit as blackbodies, (2) No attenuation of solar irradiation
enroute to earth, and (3) Earth atmosphere has no effect on earth energy balance.
ANALYSIS: Performing an energy balance on the earth,
E& in − E& out = 0
Ae,p ⋅G S = Ae,s ⋅ Eb (Te )
(π D2e / 4 )GS = π D2eσ Te4
Te = ( G S / 4σ )
1/4
where Ae,p and Ae,s are the projected area and total surface area of the earth, respectively. To
determine the irradiation GS at the earth’s
surface, equate the rate of emission from the
sun to the rate at which this radiation passes
through a spherical surface of radius RS,e – De/2.
E& in − E& out = 0
2
π DS2 ⋅σ TS4 = 4π  RS,e − De / 2  GS
(
π 1.39 × 109 m
) × 5.67 ×10−8 W / m2 ⋅ K4 ( 5800 K)4
2
2
= 4π 1.5× 1011 − 1.29 × 107 / 2 m2 × GS


GS = 1377.5 W / m 2 .
Substituting numerical values, find
(
Te = 1377.5 W / m 2 / 4 × 5.67 × 10−8 W / m 2 ⋅ K4
)
1/4
= 279 K.
<
COMMENTS: (1) The average earth’s temperature is greater than 279 K since the effect of the
atmosphere is to reduce the heat loss by radiation.
(2) Note carefully the different areas used in the earth energy balance. Emission occurs from the total
spherical area, while solar irradiation is absorbed by the projected spherical area.
PROBLEM 12.58
KNOWN: Spectral transmissivity of low iron glass (see Fig. 12.24).
FIND: Interpretation of the greenhouse effect.
SCHEMATIC:
ANALYSIS: The glass affects the net radiation transfer to the contents of the greenhouse. Since
most of the solar radiation is in the spectral region λ < 3 µm, the glass will transmit a large fraction of
this radiation. However, the contents of the greenhouse, being at a comparatively low temperature,
emit most of their radiation in the medium to far infrared. This radiation is not transmitted by the glass.
Hence the glass allows short wavelength solar radiation to enter the greenhouse, but does not permit
long wavelength radiation to leave.
PROBLEM 12.71
KNOWN: Temperature, absorptivity, transmissivity, radiosity and convection conditions for a
semitransparent plate.
FIND: Plate irradiation and total hemispherical emissivity.
SCHEMATIC:
ASSUMPTIONS: (1) Steady-state conditions, (2) Uniform surface conditions.
ANALYSIS: From an energy balance on the plate
E& in = E& out
2G = 2q ′′conv + 2J.
Solving for the irradiation and substituting numerical values,
G = 40 W / m 2 ⋅ K ( 350 − 300 ) K + 5000 W / m 2 = 7000 W / m 2 .
<
From the definition of J,
J = E + ρG + τ G = E + (1 −α ) G.
Solving for the emissivity and substituting numerical values,
ε=
J − (1 − α ) G
σT
4
5000 W / m 2 ) − 0.6 ( 7000 W / m 2 )
(
=
= 0.94.
4
5.67 ×10 −8 W / m 2 ⋅ K 4 ( 350K)
<
Hence,
α ≠ε
and the surface is not gray for the prescribed conditions.
COMMENTS: The emissivity may also be determined by expressing the plate energy balance as
2α G = 2q ′′conv + 2E.
Hence
ε σ T 4 = αG − h ( T − T∞ )
ε=
(
)
0.4 7000 W / m 2 − 40 W / m 2 ⋅ K ( 50 K)
4
5.67 ×10 −8 W / m 2 ⋅ K 4 ( 350 K )
= 0.94.
PROBLEM 13.1
KNOWN: Various geometric shapes involving two areas A1 and A2.
FIND: Shape factors, F12 and F21, for each configuration.
ASSUMPTIONS: Surfaces are diffuse.
ANALYSIS: The analysis is not to make use of tables or charts. The approach involves use of the
reciprocity relation, Eq. 13.3, and summation rule, Eq. 13.4. Recognize that reciprocity applies to two
surfaces; summation applies to an enclosure. Certain shape factors will be identified by inspection.
Note L is the length normal to page.
(a) Long duct (L):
<
By inspection, F12 = 1.0
By reciprocity, F21 =
A1
A2
F12 =
2 RL
(3 / 4 ) ⋅ 2π RL
× 1.0 =
4
3π
= 0.424
<
(b) Small sphere, A1, under concentric hemisphere, A2, where A2 = 2A
Summation rule
F11 + F12 + F13 = 1
But F12 = F13 by symmetry, hence F12 = 0.50
By reciprocity,
F21 =
A1
A2
F12 =
A1
2A1
× 0.5 = 0.25.
<
<
(c) Long duct (L):
<
By reciprocity,
F21 =
A1
A2
F12 =
2RL
π RL
× 1.0 =
2
π
Summation rule,
F22 = 1 − F21 = 1 − 0.64 = 0.363.
Summation rule,
F11 + F12 + F13 = 1
= 0.637
<
<
By inspection,
F12 = 1.0
(d) Long inclined plates (L):
But F12 = F13 by symmetry, hence F12 = 0.50
By reciprocity,
(e) Sphere lying on infinite plane
Summation rule,
F21 =
A1
A2
F12 =
20L
10 ( 2 )
1/ 2
× 0.5 = 0.707.
L
F11 + F12 + F13 = 1
But F12 = F13 by symmetry, hence F12 = 0.5
By reciprocity,
<
<
F21 =
A1
A2
F12 → 0 since
A 2 → ∞.
Continued …..
<
<
PROBLEM 13.1 (Cont.)
(f) Hemisphere over a disc of diameter D/2; find also F22 and F23.
<
By inspection, F12 = 1.0
Summation rule for surface A3 is written as
F31 + F32 + F33 = 1. Hence, F32 = 1.0.
By reciprocity,
F23 =
A3
F32
A2
 2 π D / 2 2 
( )  / π D2 1.0 = 0.375.
 πD
F23 =  
−

4
 2 
  4


By reciprocity,
 π  D  2 π D2 
A1
F21 =
F12 =    /
 × 1.0 = 0.125.
A2
2 
 4  2 

<
F21 + F22 + F23 = 1 or
Summation rule for A2,
F22 = 1 − F21 − F23 = 1 − 0.125 − 0.375 = 0.5.
<
Note that by inspection you can deduce F22 = 0.5
(g) Long open channel (L):
Summation rule for A1
F11 + F12 + F13 = 0
<
but F12 = F13 by symmetry, hence F12 = 0.50.
By reciprocity,
F21 =
A1
A2
F12 =
2× L
( 2π 1) / 4 × L
=
4
π
× 0.50 = 0.637.
COMMENTS: (1) Note that the summation rule is applied to an enclosure. To complete the
enclosure, it was necessary in several cases to define a third surface which was shown by dashed
lines.
(2) Recognize that the solutions follow a systematic procedure; in many instances it is possible to
deduce a shape factor by inspection.
PROBLEM 13.7
KNOWN: Right-circular cylinder of diameter D, length L and the areas A1, A2, and A3 representing
the base, inner lateral and top surfaces, respectively.
FIND: (a) Show that the view factor between the base of the cylinder and the inner lateral surface
has the form
4
!
F12 = 2 H 1 + H 2
9
1/ 2
−H
"#
$
where H = L/D, and (b) Show that the view factor for the inner lateral surface to itself has the form
4
F22 = 1 + H − 1 + H 2
9
1/ 2
SCHEMATIC:
ASSUMPTIONS: Diffuse surfaces with uniform radiosities.
ANALYSIS: (a) Relation for F12, base-to-inner lateral surface. Apply the summation rule to A1,
noting that F11 = 0
F11 + F12 + F13 = 1
F12 = 1 − F13
(1)
From Table 13.2, Fig. 13.5, with i = 1, j = 3,
F13 =
%K& K' !
1
6 "#$
1/ 2
1
S − S2 − 4 D3 / D1 2
2
S = 1+
1 + R 23
R12
=
1
R2
(K)
K*
(2)
+ 2 = 4 H2 + 2
(3)
where R1 = R3 = R = D/2L and H = L/D. Combining Eqs. (2) and (3) with Eq. (1), find after some
manipulation
Continued …..
PROBLEM 13.7 (Cont.)
%K
4
&K
!
'
1/ 2
"
F12 = 2 H 41 + H 2 9 − H #
!
$
9 "#$
1/ 2
2
1
2
2
F12 = 1 − 4 H + 2 − 4 H + 2 − 4
2
(K
)K
*
(4)
(b) Relation for F22, inner lateral surface. Apply summation rule on A2, recognizing that F23 = F21,
F21 + F22 + F23 = 1
F22 = 1 − 2 F21
(5)
Apply reciprocity between A1 and A2,
1
6
F21 = A1 / A 2 F12
(6)
and substituting into Eq. (5), and using area expressions
F22 = 1 − 2
A1
D
1
F12 = 1 − 2
F12 = 1 −
F12
A2
4L
2H
(7)
2
where A1 = πD /4 and A2 = πDL.
Substituting from Eq. (4) for F12, find
F22 = 1 −
4
!
9
"#
$
4
9
1/ 2
1/ 2
1
2 H 1 + H2
− H = 1+ H − 1 + H2
2H
<
PROBLEM 13.19
KNOWN: Arrangement of three black surfaces with prescribed geometries and surface temperatures.
FIND: (a) View factor F13, (b) Net radiation heat transfer from A1 to A3.
SCHEMATIC:
ASSUMPTIONS: (1) Interior surfaces behave as blackbodies, (2) A2 >> A1.
ANALYSIS: (a) Define the enclosure as the interior of the cylindrical form and identify A4.
Applying the view factor summation rule, Eq. 13.4,
F11 + F12 + F13 + F14 = 1.
(1)
Note that F11 = 0 and F14 = 0. From Eq. 13.8,
F12 =
D2
D2 + 4L2
=
(3m )2
= 0.36.
2
2
(3m ) + 4 ( 2m )
(2)
From Eqs. (1) and (2),
<
F13 = 1 − F12 = 1 − 0.36 = 0.64.
(b) The net heat transfer rate from A1 to A3 follows from Eq. 13.13,
(
q13 = A1 F13 σ T14 − T34
)
(
)
q13 = 0.05m 2 × 0.64 × 5.67 ×10−8 W / m 2 ⋅ K 4 10004 − 5004 K 4 = 1700 W.
COMMENTS: Note that the summation rule, Eq. 13.4, applies to an enclosure; that is, the total
region above the surface must be considered.
<